Integration by partial fractions

  • Thread starter Zeth
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  • #1
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The problem is
[tex]\int \frac{2s+2}{(s^2+1)(s-1)^3} dx[/tex]
What I'm wondering about is there anyway to get the partial fractions out without doing the full mess of bringing up the [tex] (s^2+1) and (s-1)^3 [/tex]? I tried the heaviside method and got one of the numerators but I'm stuck for a practical way to do the others.
 

Answers and Replies

  • #2
Gib Z
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Sorry you'll have to "bring it up", but a tiny shortcut - letting s=1 should give you some insight, if not, compare co efficients.
 
  • #3
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Sorry you'll have to "bring it up", but a tiny shortcut - letting s=1 should give you some insight, if not, compare co efficients.

setting s=1 is the heaviside method
 
  • #4
Gib Z
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Lol my bad, never heard it be called that before.
 
  • #5
The problem is
[tex]\int \frac{2s+2}{(s^2+1)(s-1)^3} dx[/tex]
What I'm wondering about is there anyway to get the partial fractions out without doing the full mess of bringing up the [tex] (s^2+1) and (s-1)^3 [/tex]? I tried the heaviside method and got one of the numerators but I'm stuck for a practical way to do the others.
A nice shortcut I saw in one of my math tutes is calculating residues for each of the poles of the above function and forming the sum of rational functions whose numerators are the corresponding resides and the denominators being s - pole.
Not sure if that would work here, since we're not dealing with first order poles as we were in class. Furthermore, I'm not sure whether or not you'd like to move into the realm of complex numbers.
What's wrong with just comparing co-efficients, or using Heaviside's method?
Also, you might want to change that "dx" to a "ds", unless [tex]s=s(x)[/tex]?
 
  • #6
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What's wrong with just comparing co-efficients, or using Heaviside's method?
Also, you might want to change that "dx" to a "ds", unless [tex]s=s(x)[/tex]?

My bad, I keep putting x everywhere instead of the proper dummy variable. I'll have a look at what you said even though I did do the integral by comparing coefficients and I did use the Heaviside method to get the one numerator I could. I wanted to avoid comparing coefficients because it's very messy and some of the other problems looked worse, but weren't once I tried them.

On a related note the next problem has [tex]\int \frac{2\theta+1}{\theta^2+2\theta+2} d\theta[/tex] which I'm not sure how to do, I've tried setting [tex] u = \theta^2[/tex] and [tex] u = 2\theta [/tex] but I didn't get far. Any ideas on what to do with that one?
 
  • #7
Dick
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Complete the square in the denominator and use the thing being squared as u.
 
  • #8
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I'm probably wrong but I tried that and it doesn't seem to work:
I end up with
[tex]\int \frac{2\theta+1}{(\theta+1)^2+1} d\theta[/tex]
and
[tex]u = \theta +1 [/tex] [tex]du = d\theta[/tex]
but the top line still looks like
[tex]2\theta +1[/tex]
best I've managed is:
[tex] \int \frac {(2\theta +2)-1}{(\theta+1)^2+1} d\theta[/tex]
which I still have no idea how to do with
[tex] u = \theta +1[/tex] and [tex] du = d\theta[/tex]
[tex] \int \frac {2u-1}{u^2+1} du[/tex]
 
Last edited:
  • #9
Dick
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I'm probably wrong but I tried that and it doesn't seem to work:
I end up with
[tex]\int \frac{2\theta+1}{(\theta+1)^2+1} d\theta[/tex]
and
[tex]u = \theta +1 [/tex] [tex]du = d\theta[/tex]
but the top line still looks like
[tex]2\theta +1[/tex]
best I've managed is:
[tex] \int \frac {(2\theta +2)-1}{(\theta+1)^2+1} d\theta[/tex]
which I still have no idea how to do with
[tex] u = \theta +1[/tex] and [tex] du = d\theta[/tex]
[tex] \int \frac {2u-1}{u^2+1} du[/tex]

Now just split the integral in two. u/(u^2+1) is a simple substitution. 1/(u^2+1) ought to look familiar.
 
  • #10
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lol I can't believe I missed that, sleep it is.
Thanks a lot.
 

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